Area Under the Cycloid:
Mamikon´s Method
Tomás Garza
February 2019

Background

This notebook is based on a clip made by Ujjwal Rane using MicroStation. This is a software for CAD and I thought it might be interesting to reproduce the idea using Mathematica and check whether a similar result is obtained. It does, as seen here.

Description

Mamikon Mnatsakanian, usually known as Mamikon, invented back in 1959 an approach to solve a wide range of problems in Geometry. A description of the method may be found in Wikipedia, in the entry Visual Calculus. The method has had an enthusiastic welcome by the contemporary mathematical community; in particular, Tom A. Apostol has co-authored a number of publications with Mamikon and successfully promoted his ideas in the course of the last 30 years.

Here we give the application of the method to find the area of a cycloid, a problem which conventionally requires the use of integral calculus for its solution.

The problem

The cycloid is the path described by the extreme of a given radius of a circle rolling along a straight line. In the figure below, the green point is assumed to lie initially on the the straight line. As the circle rolls -moving the slider to the right- it will touch the line again when the radius has moved through an angle of 2 π . As the rolling proceeds, the green point describes the path in orange.

Out[]=

angle t swept by

the green radius

0.0

radians

The problem is to find the area under the orange curve, once a whole turn has been completed.

Solution: step one

First, draw a rectangle containing the orange path. See this in the figure below with t = 0.

Out[]=

angle t swept by

the green radius

0.0

radians

When the circle rolls all the way to the extreme, i.e., when t = 2 π, one can distinguish three regions: the rectangle, the area under the curve (in red), and the region (in gray) above the orange curve. Mamikon’s method obtains the area of the region in gray using the tangents to the cycloid as the circle rolls out, wherefrom the area under the curve may be obtained directly.

Solution: step two

Now, in the figure below, for each value of t, 0 ≤ t ≤ 2 π, a tangent to the cycloid is drawn (in yellow) reaching the upper side of the rectangle. The tangent originates at the green point.

Out[]=

angle t swept by

the green radius

0.0

radians

In order to have a better understanding of what goes on as the circle rolls out, we reproduce the behavior of the tangentes inside a stationary circle of the same diameter as the green one. This will appears above the rolling circle.

Solution: step three

In the figure below, each yellow segment (tangent to the cycloid) is a chord within the green circle. As t varies, these segments sweep the region between the cycloid and the rectangle. The stationary circle in the upper part reproduces what is going below: the yellow lines are precisely the same as the corresponding ones below, and they share a common vertex at the top of the circle. Incidentally, the circle at the top is known as the tangent cluster.

So, for example, when t = π, the area swept by the tangents in the lower part is half the total area, and this is reflected in the tangent cluster, where the swept area is half the circle.

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